### Example Problem **Problem:** Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive? **Explanation:** In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions. 1. **Average Rate of Customer Arrivals (λ):** - The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute. 2. **Exponential Distribution Function:** - For a continuous variable, the probability density function (PDF) for the exponential distribution is given by: \[ f(t) = λe^{-λt} \] where \( t \) is the time in minutes. 3. **Calculate the Probability:** - To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference: \[ P(2 \leq t \leq 4) = F(4) - F(2) \] where \( F(t) = 1 - e^{-λt} \). 4. **Applying the Values:** - Compute \( F(4) \) and \( F(2) \): \[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \] \[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \] - Now find the difference: \[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \] \[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \] By solving the above expression using a calculator, you would get the exact numerical likelihood value. **Note:** The example

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Question 9 Please use simple probability rules
### Example Problem

**Problem:**
Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive?

**Explanation:**
In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions. 

1. **Average Rate of Customer Arrivals (λ):** 
   - The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute.

2. **Exponential Distribution Function:**
   - For a continuous variable, the probability density function (PDF) for the exponential distribution is given by:
     \[ f(t) = λe^{-λt} \]
     where \( t \) is the time in minutes.

3. **Calculate the Probability:**
   - To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference:
     \[ P(2 \leq t \leq 4) = F(4) - F(2) \]
     where \( F(t) = 1 - e^{-λt} \).

4. **Applying the Values:**
   - Compute \( F(4) \) and \( F(2) \):
     \[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \]
     \[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \]
   - Now find the difference:
     \[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \]
     \[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \]

By solving the above expression using a calculator, you would get the exact numerical likelihood value.

**Note:** The example
Transcribed Image Text:### Example Problem **Problem:** Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive? **Explanation:** In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions. 1. **Average Rate of Customer Arrivals (λ):** - The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute. 2. **Exponential Distribution Function:** - For a continuous variable, the probability density function (PDF) for the exponential distribution is given by: \[ f(t) = λe^{-λt} \] where \( t \) is the time in minutes. 3. **Calculate the Probability:** - To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference: \[ P(2 \leq t \leq 4) = F(4) - F(2) \] where \( F(t) = 1 - e^{-λt} \). 4. **Applying the Values:** - Compute \( F(4) \) and \( F(2) \): \[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \] \[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \] - Now find the difference: \[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \] \[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \] By solving the above expression using a calculator, you would get the exact numerical likelihood value. **Note:** The example
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